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A series L-C-R circuit is connected acro...

A series L-C-R circuit is connected across an AC source E = `10sin[100pit - pi/6]`. Current from the supply is I = `2sin[100pit + pi/12]`, What is the average power dissipated?

Text Solution

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Phase difference between voltage and current
`phi = (pi)/(12) - (-(pi)/(6))=(pi)/(4)`
`rArr` Power factor ` = cos phi = (1)/(sqrt(2))`
Average power dissipated `= (V_(m)l_(m))/(2) cos phi = (10xx2)/(2) xx (1)/(sqrt(2))=5sqrt(2)` watt
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